Summary: Continuing notes on the tensor version of the Lorentz tranform. It's time to start on the second set of examples.

The interval in four space is invariant under Lorentz transforms and is called the Lorentz scalar.

The Lorentz transform also applies to differential distances as,

$dx^{\prime\mu} = \Lambda^\mu_\nu x^\mu$

We were asked in class to work out $x^2+y^2+z^2-t^2 = x^{\prime 2}+y^{\prime 2}+z^{\prime 2}-t^{\prime 2}$

The transforms we'll use are:

$x = \gamma\left(x^\prime + vt^\prime\right)$

$t = \gamma\left(t^\prime + vx^\prime\right)$

Substituting these into the l.h.s. gives

$\gamma^2\left(x^\prime + vt^\prime\right)^2 - \gamma^2\left(t^\prime + vx^\prime\right)^2 = x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} +2vtx + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + 2vxt+v^2x^{\prime 2}\right)= x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + v^2x^{\prime 2}\right) = x^{\prime 2} - t^{\prime 2}$

$\gamma^2 x^{\prime 2} - \gamma^2v^2x^{\prime 2} = \gamma^2 x^{\prime 2}\left(1 - v^2\right) = x^{\prime 2}$

Similarly

$\gamma^2 v^{\prime 2}t^{\prime 2} - \gamma^2 t^{\prime 2} = -\gamma^2 t^{\prime 2}\left(-v^2 + 1\right) = -t^{\prime 2}$

So

$x^{\prime 2} - t^{\prime 2} = x^{\prime 2} - t^{\prime 2}$

Done

## Thursday, September 4, 2014

### Showing that SpaceTime Intervals are invariant: EM II notes 2014_09_03

Posted by Hamilton Carter at 6:51 AM

Labels: emII, lorentz transform, physics, special relativity

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